Here's a parameterization of a plane: $\vec{v}(x, y) = (3x - 3y, x + 3y, 2x + y)$ What vectors are normal to the plane $\vec{v}$ ? Choose 2 answers: Choose 2 answers: (Choice A) A $(-5, -9, 12)$ (Choice B) B $(5, 9, -12)$ (Choice C) C $(12, -9, -5)$ (Choice D) D $(-12, 9, 5)$
Solution: The vector normal to the area element describes what's perpendicular to the surface, and the vector's magnitude represents the area of a tiny rectangle along the parameterization. We can take multiply the result by $-1$ to find another normal vector. $\text{normal to area element} = \dfrac{\partial \vec{v}}{\partial x} \times \dfrac{\partial \vec{v}}{\partial y}$ Let's calculate the cross product. $\begin{aligned} \dfrac{\partial \vec{v}}{\partial x} \times \dfrac{\partial \vec{v}}{\partial y} &= \det \begin{pmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ \\ 3 & 1 & 2 \\ \\ -3 & 3 & 1 \end{pmatrix} \\ \\ &= -5 \hat{\imath} - 9\hat{\jmath} + 12 \hat{k} \end{aligned}$ We want two normal vectors. Because the negative of a normal vector is also normal to the surface, we can take the negative of what we just calculated as a second normal vector to the plane $\vec{v}$. Therefore, two vectors normal to $\vec{v}$ are $(-5, -9, 12)$ and $(5, 9, -12)$.